January 25, 2008
PHY2053 Discussion
Quiz 2 (Chapter 2.5-2.6 & 3.1-3.3)
Name:
UFID:
**1. (5 pts) In an auto race, initially two cars are moving side by side at 80.0 m/s. Then
one of the cars makes a pit stop. It slows to a stop over a distance of 320 m, spends 10.0 s
in the pit and then accelerates out, reaching its initial speed 80.0 m/s after a distance of
400 m. At this point, what is the distance between the two cars?
Let the car which makes a pit stop be A and the other one be B. The distance between
the two cars is given by
ΔX = Xb – Xa = Xb -(320+400) = Xb-720 m
Since displacement is given by average velocity times time interval, the time intervals
for the car A to stop and reach its initial velocity are given by
40t = 320 ⇒t = 8 s & 40t’ = 400 ⇒ t’ = 10 s
Thus the car B travels
Xb = 80(8+10+10) = 2240 m
Therefore, ΔX = 2240-720 = 1520 m
*2. (5pts) A stone is thrown vertically upward from the ground with an initial speed of
30.0 m/s. What is the maximum height the stone reaches?
Since the velocity of the stone is zero at the top, we get
v² = v’² + 2aΔy ⇒ 0² = 30²+2(-9.8)Δy ⇒ Δy = 900/19.6 =45.9 m
***3. (5 pts) A ball is thrown upward from the ground with an initial speed of 25 m/s;
one second later, another ball is thrown downward with an initial speed of 5 m/s from
the top of a building 50 m high. Find the height of the two balls when they cross each
other.
The height of the ball thrown upward is given by
Δy = y-0 = 25t-(1/2)9.8t²
Since the other ball is thrown one second later, it falls only for t-1 second. Thus the
height of this ball is given by
Δy’ = y’-50 = -5(t-1)-(1/2)9.8(t-1)²
When they pass each other, their height is the same, thus
y = y’ ⇒ 25t-4.9t² = 50-5(t-1)-4.9(t-1)² ⇒ 20.2t = 50.1 ⇒ t = 2.48 s
The height is given by
y (=y’) = 25×2.48-4.9(2.48)² = 31.9 m
**4. (5 pts) The eye of a hurricane passes over a city with a speed of 40 km/h due
northeast (45º north of east). Three hours later, the hurricane shifts the course and its
speed, and it starts to move in a direction 30º east of north with a speed of 30 km/h. Find
the direction of the hurricane in respect to the city 6 hours after it passes over the city.
We express the displacements for the first 3 hours and the second 3 hours in
components:
Δr = (40cos45º km/h, 40sin45º km/h)×3 h = (84.9 km, 84.9 km)
Δr’ = (30cos60º, 30sin60º)×3 = (45.0 km, 77.9 km)
The total displacement is the sum of these two vectors:
Δr+Δr’ = (84.9, 84.9)+(45, 77.9) = (130 km, 163 km)
Converting the component expression to the angle, we get
θ = tanˉ¹(163/130) = 51.4º (51.4 N of E or 38.6 E of N)